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আপনার কাছে যদি মনে হয় প্রশ্নটি অধ্যায় অনুযায়ী সঠিক নয় তাহলে সঠিক অধ্যায় ও প্রশ্নটি অথবা কোন প্রকার ভুল থাকলে আমাদের কে জানান ইমেইল করে kabirdepart@gmail.com

LCM and HCF

Correct :

Wrong :

1. What is the least number when divided by 4,5,6 leaves remainder 3 in each case?
a) 33
b) 43
c) 53
d) 63
e) 72
LCM of 4, 5, 6=60

So, 60+3=63

2. What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30?
a) 196
b) 630
c) 1260
d) 2520

3. The ratio of two numbers is 3:4 and their H.C.F is 4. Their L.C.M is:
a) 48
b) 12
c) 16
d) 24
Let the numbers be 3x and 4x. Then, their H.C.F. = x. So, x = 4.

So, the numbers 12 and 16.

L.C.M. of 12 and 16 = 48.

4. The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. The sum of the numbers is :
a) 28
b) 32
c) 40
d) 64

5. What is the minimum number of chocolates that must be added to an existing stock of 966 chocolates, so that the total stock can be equally distributed among 6, 7, 8 or 9 persons?
a) 36
b) 42
c) 56
d) 72
e) none of these
prime factors of following numbers are,
6=2x3
7=1x7
8=2x2x2x2
9=3x3
so,L.C.M=2x3x3x7x4=504

6. The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is:
a) 504
b) 536
c) 544
d) 548
Required number = (L.C.M. of 12, 15, 20, 54) + 8

= 540 + 8

= 548.

7. If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:
a) 55/601
b) 601/55
c) 11/120
d) 120/11
Let the numbers be a and b.

Then, a + b = 55 and ab = 5 x 120 = 600.

The required sum = 1/a+1/b

= (a + b)/ab= 55/600 = 11/120

8. The H.C.F. of 9/10 , 12/25 , 18/35 and 21/40 is:
a) 3/5
b) 252/5
c) 3/1400
d) 63/700

9. The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is:
a) 28
b) 32
c) 40
d) 64
Let the numbers be 2x and 3x.

Then, their L.C.M. = 6x.

So, 6x = 48 or x = 8.

The numbers are 16 and 24.

Hence, required sum = (16 + 24) = 40.

10. Which of the following has the most number of divisors?
a) 99
b) 101
c) 176
d) 182
99 = 1 x 3 x 3 x 11

101 = 1 x 101

176 = 1 x 2 x 2 x 2 x 2 x 11

182 = 1 x 2 x 7 x 13

So, divisors of 99 are 1, 3, 9, 11, 33, .99

Divisors of 101 are 1 and 101

Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176

Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.

Hence, 176 has the most number of divisors.

11. The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:
a) 123
b) 127
c) 235
d) 305
Required number = H.C.F. of (1657 - 6) and (2037 - 5)

= H.C.F. of 1651 and 2032 = 127.

12. Which of the following fraction is the largest?
a) 7/8
b) 13/16
c) 31/40
d) 63/80

13. Find the highest common factor of 36 and 84.
a) 4
b) 6
c) 12
d) 18
36 = 2² x 3²

84 = 2² x 3 x 7

So, H.C.F. = 2² x 3 = 12.

14. Three numbers which are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is:
a) 75
b) 81
c) 85
d) 89
Since the numbers are co-prime, they contain only 1 as the common factor.

Also, the given two products have the middle number in common.

So, middle number = H.C.F. of 551 and 1073 = 29;

First number = (551/29) = 19;

Third number = (1073/29) = 37.

So, Required sum = (19 + 29 + 37) = 85.

15. The greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12 m 95 cm is:
a) 15 cm
b) 25 cm
c) 35 cm
d) 42 cm
Required length = H.C.F. of 700 cm,

385 cm and 1295 cm = 35 cm.

16. 252 can be expressed as a product of primes as:
a) 2 x 2 x 3 x 3 x 7
b) 2 x 2 x 2 x 3 x 7
c) 3 x 3 x 3 x 3 x 7
d) 2 x 3 x 3 x 3 x 7
Clearly, 252 = 2 x 2 x 3 x 3 x 7.

17. The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is:
a) 1008
b) 1015
c) 1022
d) 1032
Required number = (L.C.M. of 12,16, 18, 21, 28) + 7

= 1008 + 7

= 1015

18. The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is:
a) 12
b) 16
c) 24
d) 48
Let the numbers be 3x and 4x.

Then, their H.C.F. = x. So, x = 4.

So, the numbers 12 and 16.

L.C.M. of 12 and 16 = 48.

19. What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30?
a) 196
b) 630
c) 1260
d) 2520
L.C.M. of 12, 18, 21, 30

= 2 x 3 x 2 x 3 x 7 x 5 = 1260

Required number = (1260 ÷ 2)=630.

20. The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is:
a) 279
b) 283
c) 308
d) 318
Other number = (11 x 7700)/275 = 308.

21. A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the starting point?
a) 26 minutes and 18 seconds
b) 42 minutes and 36 seconds
c) 45 minutes
d) 46 minutes and 12 seconds
L.C.M. of 252, 308 and 198 = 2772.

So, A, B and C will again meet at the starting point in 2772 sec.

i.e., 46 min. 12 sec.

22. The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:
a) 1677
b) 1683
c) 2523
d) 3363
L.C.M. of 5, 6, 7, 8 = 840.

Required number is of the form 840k + 3

Least value of k for which (840k + 3) is divisible by 9 is k = 2.

Required number = (840 x 2 + 3) = 1683.

23. Reduce (128352)/(238368) to its lowest terms:
a) 3/4
b) 5/13
c) 7/13
d) 9/13

24. The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:
a) 3
b) 13
c) 23
d) 33
L.C.M. of 5, 6, 4 and 3 = 60.

On dividing 2497 by 60, the remainder is 37.

Number to be added = (60 - 37) = 23.

25. Find the lowest common multiple of 24, 36 and 40.
a) 120
b) 240
c) 360
d) 480
L.C.M. = 2 x 2 x 2 x 3 x 3 x 5 = 360.

26. The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
a) 74
b) 94
c) 184
d) 364
L.C.M. of 6, 9, 15 and 18 is 90.

Let required number be 90k + 4, which is multiple of 7.

Least value of k for which (90k + 4) is divisible by 7 is k = 4.

Required number = (90 x 4) + 4 = 364.

27. The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:
a) 1
b) 2
c) 3
d) 4
Let the numbers 13a and 13b.

Then, 13a x 13b = 2028

=> ab = 12.

Now, the co-primes with product 12 are (1, 12) and (3, 4).

[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]

So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).

Clearly, there are 2 such pairs.

28. The G.C.D. of 1.08, 0.36 and 0.9 is:
a) 0.03
b) 0.9
c) 0.18
d) 0.108
Given numbers are 1.08, 0.36 and 0.90. H.C.F. of 108, 36 and 90 is 18,

So, H.C.F. of given numbers = 0.18.

29. Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:
a) 40
b) 80
c) 120
d) 200
Let the numbers be 3x, 4x and 5x.

Then, their L.C.M. = 60x.

So, 60x = 2400 or x = 40.

The numbers are (3 x 40), (4 x 40) and (5 x 40).

Hence, required H.C.F. = 40.

30. The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:
a) 101
b) 107
c) 111
d) 185
Let the numbers be 37a and 37b.

Then, 37a x 37b = 4107

=> ab = 3.

Now, co-primes with product 3 are (1, 3).

So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).

So, Greater number = 111.

31. The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:
a) 9000
b) 9400
c) 9600
d) 9800
Greatest number of 4-digits is 9999.

L.C.M. of 15, 25, 40 and 75 is 600.

On dividing 9999 by 600, the remainder is 399.

Required number (9999 - 399) = 9600.

32. Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:
a) 4
b) 5
c) 6
d) 8
N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)

= H.C.F. of 3360, 2240 and 5600 = 1120.

Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

33. Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together?
a) 4
b) 10
c) 15
d) 16
L.C.M. of 2, 4, 6, 8, 10, 12 is 120.

So, the bells will toll together after every 120 seconds(2 minutes).

In 30 minutes, they will toll together (30/2)+1 = 16 times.

34. The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:
a) 276
b) 299
c) 322
d) 345
Clearly, the numbers are (23 x 13) and (23 x 14).

So, Larger number = (23 x 14) = 322.

35. Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
a) 4
b) 7
c) 9
d) 13
Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43)

= H.C.F. of 48, 92 and 140 = 4.

36. How many pupils can be seated in a room with N single chairs and B double seated benches?
a) NB
b) 2NB
c) 2(B + N)
d) 2B+N
e) None of these

37. Two numbers are in the ratio of 15:11. If their H.C.F. is 13, find the numbers.
a) 195 and 143
b) 190 and 143
c) 195 and 145
d) none of these
একটি সংখ্যা ১৫x ও ১১x

প্রশ্নমতে ,x =১৩

১৫×১৩=১৯৫

১১×১৩=১৪৩

38. What is the H.C.F. of the numbers 36, 54 and 90?
a) 6
b) 9
c) 12
d) 18
e) None of Them

39. কোন লঘিষ্ঠ সংখ্যার সাথে ৩ যোগ করলে যোগফল ২৪,৩৬ ও ৪৮ দ্বারা বিভাজ্য হবে ?
a) 89
b) 141
c) 248
d) 170
২৪= ১২ * ২ = ২*২*৩*২

৩৬= ১২*৩ = ২*২*৩*৩

৪৮ = ১২*৪ = ২*২*৩*২*২

সুতরাং, ল.সা.গু = ২*২*২*২*৩*৩ = ১৪৪

সংখ্যাটি হবে = ১৪৪ - ৩ = ১৪১

40. On sports day, if 30 children were made to stand in a column, 16 columns could be formed. If 24 children were made a stand in a column, how many columns could be formed?
a) 40
b) 20
c) 22
d) 30
e) 25

41. What is the smallest number exactly divisible by each of 12, 15, 20 and 27?
a) 360
b) 480
c) 520
d) 540
e) 820

42. The difference between two number is 5 and the difference between their squares 65. What is the larger number?
a) 13
b) 11
c) 8
d) 9

43. The average of smallest and largest primes between 60 and 80 is -
a) 60
b) 70
c) 60
d) 77

44. একটি সংখ্যা ৭৪২ থেকে যত বড় ৮৩০ থেকে তত ছোট। সংখ্যাটি কত?
a) ৭৮২
b) ৭৯০
c) ৭৮০
d) ৭৮৬

45. ৩০ থেকে ৮০ এর মধ্যবর্তী ক্ষুদ্রত্তম ও বৃহত্তম মৌলিক সংখ্যার ব্যবধান কত?
a) ৩৫
b) ৪২
c) ৪৮
d) ৫৫

46. Find the greatest number that will divide 43, 91 and 183 so as leave the same remainder in each case.
a) 4
b) 7
c) 9
d) 13

47. The sum of three consecutive multiples of 3 is 72. What is the largest number?
a) 21
b) 24
c) 27
d) 36

48. The H.C.F of two numbers is 24. The number which can be their L.C.M is ---
a) 84
b) 128
c) 148
d) 120

49. The H.C.F of two numbers is 24. The number which can be their L.C.M is-
a) 84
b) 128
c) 148
d) 120

50. If the number 481*673 is completely divisible by 9, then the smallest whole number in place of * will be:
a) 2
b) 5
c) 6
d) 7

51. The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. The sum of the numbers is:
a) 28
b) 40
c) 32
d) 64

52. ৫ এবং ৯৫ এর মধ্যে ৫ ও ৩ দ্বারা বিভাজ্য সংখ্যা মােট কয়টি?
a) ৫টি
b) ৬টি
c) ৯টি
d) ৭টি
L.C.M of 5 & 3 = 15

Number divisible by 15 = (95-5)/15 = 90/15 = 6

53. A number when divided by 627 leaves a remainder 43. By dividing the same number by 19, the remainder will be-
a) 32
b) 13
c) 43
d) 5
627 is a multiple of 19. So, if a number divided by 627 leaves a remainder 43, then the same number divided by 19 will leave the same remainder as 43 divided by 19 leaves.
43 = 19×2 + 5.
So, the remainder will be 5.

54. সবচেয়ে বড় সংখ্যা কোনটি ?
a) ৮/১২
b) ১৪/১৮
c) ২/৩
d) ১৫/২০

55. The difference of two numbers is 1365. On dividing the larger number by the smaller, we get 6 as quotient and the 15 as remainder. What is the smaller number?
a) 270
b) 1270
c) 350
d) 720
Let,
Smaller number = x,
difference =1365
So, the larger number = (x +1365)

ATQ,
6x +15 = x+ 1365
Or, 5x = 1350
Or, x= 270

56. The greatest number which will divide: 4003, 4126 and 4249, leaving the same remainder in each case:
a) 50
b) 45
c) 41
d) 43
Greatest number with which if we divide P,Q,R and it leaves same remainder in each case. Number is of form = HCF of (P-Q),(P-R).

Therefore, HCF of (4126 – 4003), (4249 – 4003) = HCF of 123, 246 = 41.

57. A number is as much greater than 36 as is less than 86. Find the number.
a) 73
b) 61
c) 43
d) 68
Let x be the required number.

Given,x is as much greater than 36 as is less than 86.

x-36 = 86-x

x+x = 86+36

2x = 122

x = 122/2 = 61

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LCM and HCF

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